# Concave mirror Convex mirror - Concave mirror Convex mirror. A real image is formed by the intersection

• View
0

0

Embed Size (px)

### Text of Concave mirror Convex mirror - Concave mirror Convex mirror. A real image is formed by the...

• Reflection in curved Mirrors

• Concave mirror Convex mirror

• A real image is formed by the intersection of real light rays. The image can be projected on a screen.

• Principal Optical AxisC

The Center of Curvature (C) is the center of the sphere from which the mirror is made.

V

The exact center of the mirror is called the Vertex point (V) The distance C-V is called: R (radius)

R

• Principal Optical Axis

F

In a concave mirror, parallel light rays get converged (= focused) upon reflection.

C

The reflected light rays intersect at a focal point F, in front of the mirror. Distance F-V is called focal distance: f.

V

R

f Rule: R = 2f

• Principal Axis

F C

The reflected light rays seem to come from a focal point F, behind the mirror. ( = a virtual focal point)

In a convex mirror, parallel light rays get diverged upon reflection.

Rule: R = 2f

V

• FC

Ray 3

Ray 1: From end of object parallel to the PA reflects through the F.

Ray 2: From end of object through F reflects parallel to PA Ray 3: From end of object through C reflects back on itself

Ray 2

Ray 1

Object

Image

The image is found where the reflected rays intersect.

• F C

Ray 3

Ray 1: From end of object parallel to PA reflects away coming from virtual F.

Ray 2: From end of object through F reflects parallel to PA Ray 3: From end of object through C reflects back on itself

Ray 2

Ray 1Object

Image

The image is found where the reflected rays intersect.

• p = + for real objects

q = + for real images (in front of mirror)

q = – for virtual images (behind mirror)

f = + real f, for a concave mirror

f = – virtual f, for a convex mirror

h and h’ = + above the PA

h and h’ = – below the PA

p q

q

+ distance – distance

In front of mirror Behind mirror

h’ = –

h = +

+ height

– height

• The location and type of an image in a curved mirror can be predicted with:

Where: f = focal distance p = object distance q = image distance

The focal distance (f) is found by:

1

f = 1

p + 1

q

f = 𝑝𝑞

(𝑝 + 𝑞)

• The magnification (= size) of an image and its attitude in a curved mirror can be found with:

Where: M = magnification level h’ = image height = hi = i q = image distance h = object height = ho = o p = object distance

M = h′

h = −

q

p

• q = + for real images (in front of mirror)

q = – for virtual images (behind mirror)

Smaller image: 0 < M < 1 Inverted image: M = –

Larger image: M > 1 Upright image: M = +

f = + (> 0) real f, (in front of) concave mirror

f = – (< 0) virtual f, (behind) convex mirror

• A concave shaving mirror has a focal length of 33 cm. Calculate

the image position of a cologne bottle placed in front of the

mirror at a distance of 93 cm. Calculate the magnification of the

image. Is the image real or virtual? Is the image inverted or

upright?

Solution 1) Given: f = + 33 cm (Concave)

p = + 93 cm

2) Asked: q? M? Attitude? Type?

3) Formula: 1

f =

1

p +

1

q and M = −

q

p

4) All units in cm, OK. Answer will be in cm.

• Solution 1) Given: f = + 33 cm (Concave)

p = + 93 cm

5) Solve: 1

f =

1

p +

1

q =

1

33 =

1

93 +

1

𝑞 =

1

𝑞 =

1

33 −

1

93 = 0.0196

𝑞 = 1

0.0196 = +51 𝑐𝑚 (+ = real image)

M = − 𝑞

𝑝 = −

51

93 = −0.55 ( – = inverted image)

• The image of a crayon appears to be 23.0 cm behind the surface

of a convex mirror and is 1.70 cm tall. If the mirror’s focal length

is 46.0 cm, how far in front of the mirror is the crayon positioned?

What is the magnification of the image? Is the image virtual or

real? Is the image inverted or upright? How tall is the actual

crayon?

Solution 1) Given: q = – 23 cm (virtual image)

h’ = +1.70 cm

f = – 46.0 cm (convex)

2) Asked: p? M? Attitude? Type? h?

3) Formula: 1

f =

1

p +

1

q and M = −

q

p =

ℎ′

4) All units in cm, OK. Answer will be in cm.

• Solution 1) Given: q = – 23 cm (virtual image)

h’ = +1.70 cm

f = – 46.0 cm (convex)

5) Solve: 1

f =

1

p +

1

q =

1

−46.0 =

1

𝑝 +

1

−23 =

1

𝑝 = −

1

46.0 +

1

23 = 0.0217

p = 1

0.0217 = +46.0 𝑐𝑚

M = − 𝑞

𝑝 = −

−23

+46 = +0.5

Recommended

Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents
Documents